We have an inequality such that $\int ^{\alpha }_{0}\sqrt {1+\cos ^{2}\left( t\right) }dt \geq \sqrt {\alpha^{2}+\sin ^{2}\alpha} $ over $ 0\leq \alpha \leq \dfrac {\pi }{2} $.
I have already proven it by considerations on the arc length of $x\to \sin(x)$, but I still wonder if there are another methods.
For any $\alpha\in I=\left(0,\frac{\pi}{2}\right]$ the inequality $$ \int_{0}^{\alpha}\sqrt{1+\cos^2 t}\,dt \geq \sqrt{\alpha^2+\sin^2\alpha} $$ is trivial since the LHS is the arc length of $x\to \sin(x)$ over $J=(0,\alpha)\subset I$ and such function is concave over $I$. On the other hand, it is also pretty accurate: the difference $E(\alpha)$ between the LHS and the RHS is increasing over $I$, attaining its maximum at $\alpha=\frac{\pi}{2}$: $$ E\left(\tfrac{\pi}{2}\right) = 0.0480030054\ldots$$ It can be improved through the same principle: for instance $$ \int_{0}^{\alpha}\sqrt{1+\cos^2 t}\,dt \geq\sqrt{\tfrac{1}{4}\alpha^2+\sin^2\tfrac{\alpha}{2}}+\sqrt{\tfrac{1}{4}\alpha^2+\left(\sin\alpha-\sin\tfrac{\alpha}{2}\right)^2}$$ has a maximum error equal to $0.0150532\ldots$